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kipiarov [429]
2 years ago
5

Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,

and released. The approximation sin θ = θ holds for all three angles, with θ3 > θ2 > θ1. How do the angular frequencies of the three pendulums compare?
Physics
1 answer:
seropon [69]2 years ago
6 0

Answer:

The angular frequencies of all the 3 pendulums shall be same.

Explanation:

The time period of a simple pendulum with the approximation sin(\theta )\approx \thetais given by:

T=2\pi\sqrt{\frac{l}{g}}

The angular frequency \omega is given by

\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}

As we can see that the angular frequency is independent on the initial angle (valid strictly for small angle approximations) we conclude that the angular frequencies of the 3 pendulums are the same.

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What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?
slavikrds [6]

I will post my work, but is that 99 degrees Celsius and 25 degrees Celsius?


All you have to do is plug in the initial temperature for gold where it says Tg and the initial temperature for the water where it says Tw and then plug that in and you will have your answer.

8 0
2 years ago
Four different observers are standing in a straight line on a street and hear a siren from a police car. Each person recorded th
Amanda [17]

Answer:

Wycleff is on block 1, Lilly is on block 4, Emilia is on block 12, and Quincy is on block 17.

Explanation:

Wycleff was at block 1 and heard a low pitch sound the whole time, so the police car must have been moving away from him.

Lilly observed was in block 4 change in pitch first.  So the car must have passed her first.

Emilia was at block 12 observed a Doppler effect after Lilly.  So the car passed her after passing Lilly

Quincy was at block 17 so she heard a high pitch sound the whole time, so the police car must have been moving toward him.

7 0
2 years ago
Construction of a solar power plant is proposed for a desert area near a school. A student has hypothesized that the shade cast
hjlf

Answer:

4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period

Explanation:

The creosote bush depends on sunlight to produce the food they require through photosynthesis. The shade from the solar panels would reduce the amount of sunlight that the bush receives. This would increase the mortality of the bush.

In order to test the hypothesis the student must record the direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period. If the plants receive sunlight less than the above amount the plants should start dying. If not then the hypothesis is false.

Hence, the answer is 4. The direct sunlight received by creosote bush in the desert area (in kWh/m2) during a 12 month period.

4 0
2 years ago
1.5 kg of air within a piston-cylinder assembly executes a Carnot power cycle with maximum and minimum temperatures of 800 K and
liraira [26]

Answer:

Explanation:

Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .

So the given process of isothermal compression must have been done at the temperature of 300K  , keeping the temperature constant .

Work done on gas at isothermal compression is equal to heat transfer .

work done on gas = 80 x 10³ J

work done on gas = n RT ln v₁ / v₂

n is number of moles v₁ and v₂ are initial and final volume

molecular weight of gas = 28.97 g

1.5 kg = 1500 / 28.97 moles

= 51.77 moles

work done on gas = n RT ln v₁ / v₂

Putting the values in the equation above

80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2

ln v₁ / .2 = .62

v₁ / .2 = 1.8589

v₁ = 0.37 m³

3 0
2 years ago
When kicking a football, the kicker rotates his leg about the hip joint. (a) If the velocity of the tip of the kicker’s shoe is
Maurinko [17]

Answer:

Part a)

\omega = 33.33 rad/s

Part b)

F = 500 N

Part c)

R_{max} = 40.8 m

Explanation:

Part a)

As we know that the velocity of the tip of the kicker's shoe is given as

v = 35 m/s

also the length of the tip of the shoe from his hip joint is given as

L = 1.05 m

now the angular speed is given as

\omega = \frac{v}{L}

\omega = \frac{35}{1.05}

\omega = 33.33 rad/s

Part b)

As we know that force on the ball is given as rate of change in momentum of the ball

so it is given as

F = \frac{\Delta P}{\Delta t}

so we have

F = \frac{m(v_f - v_i)}{\Delta t}

F = \frac{0.500(20 - 0)}{20 \times 10^{-3}}

F = 500 N

Part c)

As we know that the formula of range is given as

R = \frac{v^2 sin(2\theta)}{g}

now for maximum range we know

\theta = 45 degree

R_{max} = \frac{v^2}{g}

R_{max} = \frac{20^2}{9.8}

R_{max} = 40.8 m

6 0
2 years ago
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